The $n\text{th}$ partial sum of the series $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ is given by ${{S}_{n}}=6-\frac{3}{{{n}^{2}}}$. ${{a}_{4}}=$
Explanation: ${{a}_{4}}={{S}_{4}}-{{S}_{3}}$ $\begin{aligned} {{a}_{4}}&=\left(6-\dfrac{3}{{{4}^{2}}}\right)-\left(6-\dfrac{3}{{{3}^{2}}}\right) \\\\ &=\dfrac{1}{3}-\dfrac{3}{16} \\\\ &=\dfrac{16}{48}-\dfrac{9}{48} \\\\ &=\dfrac{7}{48} \end{aligned}$